Is there a construction and/or term for the following 'sandwich' measurement?

Suppose we have two projective measurements with elements $$E_i$$, $$i=1...m$$, and $$F_j$$, $$j=1...n$$. So we know $$F_j^2=F_j$$ and $$E_i^2=E_i$$ and $$\sum_i E_i = \sum_j F_j = \mathbb{I}$$. Then it is easy to see that the elements $$E_i F_j E_i$$ form a POVM, i.e. $$E_i F_j E_i\geq 0$$ and $$\sum_{ij} E_i F_j E_i = \mathbb{I}$$. I could imagine that such a composition of measurements appears in several different contexts and I wonder if there is a name for it? Also if it is easy to implement the measurements $$E_i$$ and $$F_j$$, then is there a way to do the combined measurement (e.g. some circuit other than just the general constructions that work for every POVM)?

Edit: The post-measurement state for the outcome $$(i,j)$$ should be $$|\psi'\rangle \propto \sqrt{E_i F_j E_i}|\psi\rangle$$ up to some unitary degree of freedom, see this wikipedia article.

• Isn't this just equivalent to measuring $E$ first and then measuring $F$, when we assume the standard projective measurement update rule? If you wanted to extend it to POVM's you could just change it to $\sqrt{E_i}F_j\sqrt{E_i}$ and you'd have the same interpretation as a sequential measurement provided you use the Lüder's update rule. Mar 2, 2022 at 14:04
• @Rammus I don't think it is the same, because the two measurements might not commute. Mar 2, 2022 at 14:08
• I might be being very stupid but does that even matter here? I've defined an ordering for the measurements, I'm merely asking what's the probability I get outcome $i$ for measurement 1 and outcome $j$ for measurement 2. This should then be given by $$\langle \psi| E_i F_j E_i |\psi\rangle.$$ If you wanted the probability for if you measure $F$ first and then $E$ you could measure $F_jE_iF_j$. I'm not asking that the properties specified by $E$ and $F$ are jointly measureable, which indeed would be an issue if the measurements do not commute. Mar 2, 2022 at 14:13
• @Rammus Hm, yes, that measurement gives the same probabilities for the outcomes, but it does not give the same post-measurement state? Mar 2, 2022 at 14:20
• No I mean adjoint (Hermitian conjugate / dagger). Sure, if you want to define the state update in another way then it no longer works. But my comments were to tell you that there is a sense in which it represents a sequential measurement. Mar 2, 2022 at 15:08

I'd say that the POVM with elements $$\mu\equiv \{E_i F_j E_i\}_{i,j}$$, where $$\mu_E\equiv \{E_i\}_i$$ and $$\mu_F\equiv\{F_j\}_j$$ are both projective POVMs, can be understood as a "fine-graining" of the POVM $$\mu_E$$.
More specifically, $$E_i F_j E_i$$ is the projector onto the linear space $$\operatorname{supp}(E_i)\cap\operatorname{supp}(F_j)$$, where $$\operatorname{supp}(E_i)\equiv I-\operatorname{ker}(E_i)$$ is the support of the projector $$E_i$$. To see this, just observe thta $$E_i F_j E_i$$ acts as the identity on any $$x\in\operatorname{supp}(E_i)\cap\operatorname{supp}(F_j)$$, and sends to zero any vector not in this subspace.
As a simple example, consider a three-dimensional space, and POVMs $$\mu_E\equiv \{|+\rangle\!\langle+|, I_3 - |+\rangle\!\langle+|\}, \qquad \mu_F \equiv \{|0\rangle\!\langle0|, |1\rangle\!\langle1|, |2\rangle\!\langle2|\},$$ with $$|+\rangle\equiv\frac1{\sqrt2}(|0\rangle+|1\rangle)$$. Then, computing $$E_i F_j E_i$$ for all $$i,j$$, and summing together identical projections arising from this, we find the associated POVM to be $$\mu = \{|+\rangle\!\langle+|, |-\rangle\!\langle-|, |2\rangle\!\langle2|\}.$$